Hi ther i need help can u help me

Can somebody help me with this question, please? I have no idea how to do this or find the answer. :/ How many ways are there to

san4es73 [151]

Answer:

28 different ways

Step-by-step explanation:

This is a combination question. Combination has to do with selection.

Total number of element in the set = 8

Number divisible by 6 or 8 are {-98, -48, -42, -36, -18, -6}

Total number divisible by 6 or 8 is 6

The number of ways we can choose 6 items from 8 is expressed as;

8C6 = 8!/(8-6)!6!

8C6 = 8!/(2)!6!

8C6 = 8*7*6!/2!6!

8C6 = 8*7/2

8C6 = 56/2

8C6 = 28 ways

Hence there are 28 different ways

5 0

2 years ago

The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell

emmasim [6.3K]

Answer:

a)

Highest (-3,-3)

Lowest (2,2)

b)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

$\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12$

completing the squares:

$\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2$

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of

$\bf f(x,y)=x^2+y^2$

subject to the constraint

$\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0$

Now we have

$\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)$

Let $\bf \lambda$ be the Lagrange multiplier.

The maximum and minimum must occur at points where

$\bf \nabla f=\lambda\nabla g$

that is,

$\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)$

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

$\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y$

Replacing in the constraint

$\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2$

from this we get

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

b)

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0

3 years ago

Vietnamese: Phát biểu định nghĩa hàm số liên tục. Khảo sát tính liên tục của hàm số sau:

aleksandr82 [10.1K]

Answer:

litrally I don't understand what you are telling

7 0

2 years ago

Quantos que é -2 ao quadrado

Olegator [25]

No estoy segura e -2 ao quadrado esa es toda la pregunta

5 0

3 years ago

A refund of $274 dollars is apparently the answer... please help!

Westkost [7]

Taxable income = 3854
they have been taking taxes out in the amount of 344 for 12 months.....344* 12 = 4128....so they took out too much....so there is a refund of 4128 - 3854 =
274 refund

6 0

3 years ago